How to expand a shell variable in a Perl command-CodePudding

I want to replace the string using a shell variable.

today =`date ' %Y%m%d'`

perl -p -i -e 's/file"20221212"/file="${today}"/g'

The expectation is file="20221215". But it failed, result is file="".

How to escape this case？

CodePudding user response：

Your issue is shell quoting. Using single quotes, ', to delimit the Perl code disables all variable expansion. You need to use double quotes, ", to get variable expansion.

Using shell without any Perl to illustrate the issue

today=`date ' %Y%m%d'`
echo 'today is $today'

will output this -- with no expansion of $today

today is $today

now with double quotes

today=`date ' %Y%m%d'`
echo "today is $today"

outputs this -- $today has been expanded.

today is 20221215

Applying that to your code (I've removed the -i option to make it easier to see the results) and escaped all double quotes in the perl code with \"

echo 'file"20221212"' >input.txt
perl -p  -e "s/file\"20221212\"/file=\"${today}\"/g" input.txt

gives

file="20221215"

CodePudding user response：

Here I am giving you the hint how it should be:

I cannot reproduce the issue from your code.

#!/bin/sh
today=`date ' %Y%m%d'`
echo "today:$today"

file="input.txt";

perl -pi -e 's/file20221212/file='${today}'/g' $file

where input.txt contains:

file20221212